ISC Class 12 Physics Board Exam Question Paper 2019 with Solutions

ISC Class 12 Physics Board Exam Question Paper with Solutions

SECTION A

1. A. Choose the correct alternative (a), (b), (c) and (d) for each of the question given below: [5 Marks]
(i) A closed surface in vacuum enclose charges –q and +3q. The total electric flux emerging out of the surface is [1 Mark]
(a) Zero
(b) \(\frac{2q}{\varepsilon_{0}}\)
(c) \(\frac{3q}{\varepsilon_{0}}\)
(d) \(\frac{4q}{\varepsilon_{0}}\)

Answer: (b) \(\frac{2q}{\varepsilon_{0}}\)

By Gauss's law, total electric flux \(\phi = \frac{q_{\text{enclosed}}}{\varepsilon_{0}} = \frac{-q + 3q}{\varepsilon_{0}} = \frac{2q}{\varepsilon_{0}}\).

Teacher's Note:
a) Apply Gauss's law taking the algebraic sum of all enclosed charges.
b) Students often forget to account for the sign of negative charges, leading to calculation errors.

(ii) What is the angle of dip at a place where the horizontal component B_H and the vertical component B_V of the earth's magnetic field are equal to [1 Mark]
(a) \(130^{\circ}\)
(b) \(60^{\circ}\)
(c) \(45^{\circ}\)
(d) \(90^{\circ}\)

Answer: (c) \(45^{\circ}\)

Since \(\tan\theta = \frac{B_V}{B_H}\) and \(B_V = B_H\), we get \(\tan\theta = 1\), which gives \(\theta = 45^{\circ}\).

Teacher's Note:
a) The angle of dip is defined as the angle made by the total magnetic field vector of the earth with the horizontal direction.
b) Note that option (a) has a printing error in the degree symbol format in the original paper, but option (c) is unambiguously correct.

(iii) A beam of light is incident at the polarizing angle of \(35^{\circ}\) on a certain glass plate. The refractive index of the glass plate is [1 Mark]
(a) \(\sin 35^{\circ}\)
(b) \(\tan 35^{\circ}\)
(c) \(\sin 55^{\circ}\)
(d) \(\tan 55^{\circ}\)

Answer: (b) \(\tan 35^{\circ}\)

According to Brewster's law, the refractive index \(n = \tan i_p\), where \(i_p\) is the polarizing angle.

Teacher's Note:
a) Recall Brewster's law relating refractive index to the tangent of the polarizing angle.
b) Do not confuse polarizing angle with critical angle or angle of refraction.

(iv) In a gamma ray emission from nucleus [1 Mark]
(a) Only the number of proton change
(b) The number of proton and neutron both changes
(c) There is no change in the number of protons and the number of neutrons
(d) Only the number of neutrons changes

Answer: (c) There is no change in the number of protons and the number of neutrons

Gamma ray emission involves the transition of a nucleus from a higher energy state to a lower energy state, releasing energy in the form of electromagnetic radiation without changing atomic or mass numbers.

Teacher's Note:
a) Gamma rays are high-frequency photons emitted during radioactive de-excitation.
b) Unlike alpha or beta decay, gamma decay does not alter the nucleon composition of the nucleus.

(v) The energy associated with light of which of the following colors in minimum? [1 Mark]
(a) Violet
(b) Red
(c) Green
(d) Yellow

Answer: (b) Red

[Figure: Multiple choice options listing colors with varying wavelengths]

Energy \(E = \frac{hc}{\lambda}\). Since red light has the maximum wavelength among visible colors, its associated energy is minimum.

Teacher's Note:
a) Energy is inversely proportional to wavelength and directly proportional to frequency.
b) Red light has the longest wavelength and lowest frequency in the visible spectrum.

B. Answer the following questions briefly and to the point.

(i) Define equipotential surface [1 Mark]

Answer:
An equipotential surface is defined as any surface that has the same electrostatic potential at every point on it.

Teacher's Note:
a) No work is required to move a test charge between two points on an equipotential surface.
b) Electric field lines are always normal to the equipotential surface at every point.

(ii) Calculate the net emf across A and B shown in the figure below [1 Mark]
[Figure: Circuit diagram showing two 4V cells connected in series in one branch, combined in parallel with an 8V cell across terminals A and B]

Answer:
Here, the two 4V cells are connected in series, so their combined emf is \(4\text{V} + 4\text{V} = 8\text{V}\). This combination is connected in parallel with another 8V cell, so the net equivalent emf across terminals A and B is \(8\text{V}\).

Teacher's Note:
a) Emf in series adds up when connected with matching polarity.
b) Parallel combination of identical emfs maintains the same terminal voltage.

(iii) Why are the pole pieces of a horseshoe magnet in moving coil galvanometer made cylindrical in shape? [1 Mark]

Answer:
The pole pieces of a horseshoe magnet in a moving coil galvanometer are made cylindrical in shape to produce a radial magnetic field, which ensures that the deflecting torque is independent of the coil's orientation.

Teacher's Note:
a) A radial magnetic field keeps the magnetic field vector perpendicular to the area vector of the coil at all positions.
b) This results in a linear scale for the galvanometer.

(iv) What is the value of power factor for pure resistor connected to an alternating current sources? [1 Mark]

Answer:
The power factor for a pure resistor connected to an alternating current source is \(1\).

Teacher's Note:
a) For a pure resistor, the phase difference \(\phi\) between alternating voltage and current is zero.
b) Power factor = \(\cos\phi = \cos 0^{\circ} = 1\).

(v) What should be the path difference between two waves reaching a point for obtaining constructive interference in Young's double slit experiment? [1 Mark]

Answer:
The path difference between two waves for obtaining constructive interference must be an integral multiple of the wavelength \(\lambda\), given by \(\Delta x = m\lambda\) (where \(m = 0, 1, 2, 3, \dots\)).

Teacher's Note:
a) Constructive interference occurs when waves arrive in phase, requiring path difference to be \(n\lambda\).
b) Mentioning the integer condition explicitly is essential for full marks.

(vi) Define the critical angle for the given medium. [1 Mark]

Answer:
Critical angle is defined as the angle of incidence in the denser medium for which the angle of refraction in the rarer medium becomes \(90^{\circ}\).

Teacher's Note:
a) Critical angle applies only when light travels from an optically denser medium to a rarer medium.
b) State both the denser medium condition and the \(90^{\circ}\) refraction angle clearly.

(vii) Name the series in the atomic spectra of the hydrogen atom that falls in the ultraviolet region. [1 Mark]

Answer:
Lyman series.

Teacher's Note:
a) Lyman series corresponds to electronic transitions to the ground state (\(n = 1\)).
b) Other series like Balmer fall in the visible region, while Paschen, Brackett, and Pfund fall in the infrared region.

SECTION B

2. In a potentiometer experiment, the balancing length with a resistance of \(2\Omega\) is found to be 100cm, while that of unknown resistance is 500cm. calculate the value of the unknown resistance. [2 Marks]

Answer:
Given balancing length \(l_1 = 100\text{ cm}\) for resistance \(R = 2\Omega\).
Balancing length for unknown resistance \(X\) is \(l_2 = 500\text{ cm}$.
Using the potentiometer relation:
\(\frac{X}{R} = \frac{l_2}{l_1}\)
\(X = \frac{l_2}{l_1} \times R = \frac{500}{100} \times 2 = 5 \times 2 = 10\Omega\).
Hence, the value of the unknown resistance is \(10\Omega\).

Teacher's Note:
a) The balancing length in a potentiometer is directly proportional to the resistance connected in the secondary circuit (\(l \propto R\)).
b) Ensure units of length are in the same ratio so conversion is not strictly required as long as both are in cm.

3. A rectangular loop of area \(5\text{ m}^{2}\), has 50 turns and carries a current of 1A. It is held in a uniform magnetic field of 0.1T, at an angle of \(30^{\circ}\). Calculate the torque experienced by the coil. [2 Marks]

Answer:
Given area \(A = 5\text{ m}^{2}\), number of turns \(N = 50\), current \(I = 1\text{ A}\), magnetic field \(B = 0.1\text{ T}\), and angle between normal to the coil and field \(\theta = 30^{\circ}\) (or angle with plane is \(30^{\circ}\), leading to angle with normal as \(90^{\circ} - 30^{\circ} = 60^{\circ}\)).
Torque \(\tau = NIAB \sin(90^{\circ} - 30^{\circ}) = NIAB \sin 60^{\circ}\)
\(\tau = 50 \times 1 \times 5 \times 0.1 \times \frac{\sqrt{3}}{2} = 25 \times \frac{\sqrt{3}}{2} = 12.5\sqrt{3}\text{ N}\cdot\text{m}\).

Teacher's Note:
a) Use the standard torque formula for a current-carrying coil: \(\tau = NIAB \sin\theta\), where \(\theta\) is the angle between the magnetic field and the normal to the plane of the coil.
b) Pay close attention to whether the given angle is with the plane of the coil or with the normal.

4. An electric current I flows through an infinitely long conductor as shown in the figure given below. Write an expression and direction for the magnetic field at point P. [2 Marks]
[Figure: Diagram showing a straight vertical long wire carrying upward current I, with point P lying at a perpendicular distance r from the wire, and magnetic field vector tangent to the circular field line]

Answer:
Magnitude of magnetic field at point P: \(B = \frac{\mu_{0}I}{2\pi r}\).
Direction: Directed tangentially along a circle of radius \(r\) centered on the conductor, pointing outward (perpendicular to the plane of paper towards the reader according to the right-hand thumb rule).

Teacher's Note:
a) Apply Ampere's Circuital Law or the Biot-Savart Law for an infinite straight wire.
b) Always state both the magnitude formula and the directional rule clearly.

5. A transformer is used to step up an alternating emf of \(200\text{V}\) to \(440\text{V}\). if the primary coil has 1000 turns, calculate the number of turns in secondary coil. [2 Marks]

Answer:
Given primary voltage \(V_P = 200\text{ V}\), secondary voltage \(V_S = 440\text{ V}\), primary turns \(N_P = 1000\).
Using the transformer ratio:
\(\frac{V_S}{V_P} = \frac{N_S}{N_P}\)
\(N_S = \frac{V_S}{V_P} \times N_P = \frac{440}{200} \times 1000 = 2.2 \times 1000 = 2200\text{ turns}\).

Teacher's Note:
a) The voltage ratio in an ideal transformer is directly proportional to the turns ratio.
b) A step-up transformer has more turns in the secondary coil than in the primary coil (\(N_S \gt N_P\)).

6. State any two properties of microwaves. [2 Marks]

Answer:
1. They can undergo reflection from metallic surfaces.
2. They exhibit polarization properties similar to light waves.

Teacher's Note:
a) Microwaves are electromagnetic waves with wavelengths ranging from \(1\text{ mm}\) to \(0.3\text{ m}\).
b) They are widely used in radar systems and microwave ovens.

7. Write any one use for each of the following mirrors. [2 Marks]
(i) Convex
(ii) Concave

Answer:
(i) Convex mirror: Used as rear-view mirrors in vehicles to provide a wider field of view.
(ii) Concave mirror: Used by doctors (ophthalmoscopes/ENT specialists) to focus light onto specific body parts.

Teacher's Note:
a) Convex mirrors always form virtual, erect, and diminished images, making them ideal as driving mirrors.
b) Concave mirrors can form real and magnified images when objects are placed close to the focus.

8. The deviation produced for violet, yellow and lights for crown glass are \(3.75^{\circ}\), \(3.25^{\circ}\) and \(2.86^{\circ}\) respectively. Calculate the dispersive power of the crown glass. [2 Marks]

Answer:
Given deviations: \(\delta_v = 3.75^{\circ}\), \(\delta_y = 3.25^{\circ}\), \(\delta_r = 2.86^{\circ}\).
Dispersive power \(\omega = \frac{\delta_v - \delta_r}{\delta_y}\)
\(\omega = \frac{3.75^{\circ} - 2.86^{\circ}}{3.25^{\circ}} = \frac{0.89^{\circ}}{3.25^{\circ}} \approx 0.274\).

Teacher's Note:
a) Dispersive power is defined as the ratio of angular dispersion to the mean deviation.
b) It is a dimensionless quantity characteristic of the prism material.

9. (i) What is the meant by the mass defect? [1 Mark]
(ii) What conclusion is drawn from Rutherford's scattering experiment of \(\alpha\text{-particles}\)? [1 Mark]

Answer:
(i) Mass defect is defined as the difference between the rest mass of the nucleus and the sum of the masses of its constituent nucleons (protons and neutrons).
(ii) It was concluded that an atom consists of a tiny, dense, positively charged center called the nucleus, and that most of the space inside the atom is empty.

Teacher's Note:
a) Mass defect is responsible for the binding energy of the nucleus via Einstein's mass-energy equivalence relation (\(E = \Delta m c^2\)).
b) The large-angle scattering of alpha particles proved the concentration of positive charge in a very small volume.

10. Define the following with reference to photoelectric effect. [2 Marks]
(i) Threshold frequency \(f_0\)
(ii) Stopping potential \(V_s\)

Answer:
(i) Threshold frequency (\(f_0\)): It is the minimum frequency of incident radiation below which no photoelectrons are emitted from a given metal surface.
(ii) Stopping potential (\(V_s\)): It is the minimum negative potential applied to the anode at which the photoelectric current becomes zero.

Teacher's Note:
a) Threshold frequency depends solely on the nature of the emitting metal surface.
b) Stopping potential measures the maximum kinetic energy of emitted photoelectrons.

11. The half-life of radium is 1550 yr. calculate its disintegration constant \(\lambda\). [2 Marks]

Answer:
Given half-life \(T_{1/2} = 1550\text{ years} = 1550 \times 3.15 \times 10^{7}\text{ s} \approx 4.88 \times 10^{10}\text{ s}\).
Using the relation for disintegration constant \(\lambda = \frac{0.6931}{T_{1/2}}\):
\(\lambda = \frac{0.6931}{1550 \times 3.15 \times 10^{7}} \approx 1.42 \times 10^{-11}\text{ s}^{-1}\) (or converted in terms of \(\text{yr}^{-1}\): \(\lambda = \frac{0.6931}{1550} \approx 4.47 \times 10^{-4}\text{ yr}^{-1}\)).

Teacher's Note:
a) Disintegration constant is inversely proportional to the half-life of a radioactive substance.
b) Pay careful attention to unit conversions between years and seconds depending on required output units.

12. Define frequency modulation and state any one advantage of frequency modulation (FM) over amplitude modulation (AM). [2 Marks]

Answer:
Frequency modulation is the process in which the frequency of the carrier wave is varied in accordance with the instantaneous amplitude of the modulating signal wave.
Advantage: FM transmission provides much better noise immunity and higher fidelity reception compared to AM.

Teacher's Note:
a) In FM, the amplitude of the carrier wave remains constant during modulation.
b) Amplitude noise does not affect frequency modulated signals, making sound quality superior.

SECTION C

13. Obtain an expression for electric potential V at the point in an end-on position, axial position of an electric dipole. [3 Marks]

Answer:
Let an electric dipole consisting of charges \(-q\) and \(+q\) separated by distance \(d\) be placed in a medium of dielectric constant \(k\). Let point P lie on the axial line at a distance \(r\) from the center of the dipole.
Distance of point P from \(+q\) is \((r - \frac{d}{2})\) and from \(-q\) is \((r + \frac{d}{2})\).
Potential due to \(+q\): \(V_1 = \frac{1}{4\pi\varepsilon_0 k} \frac{q}{r - \frac{d}{2}}\)
Potential due to \(-q\): \(V_2 = \frac{1}{4\pi\varepsilon_0 k} \frac{-q}{r + \frac{d}{2}}\)
Total potential \(V = V_1 + V_2 = \frac{q}{4\pi\varepsilon_0 k} \left[ \frac{1}{r - \frac{d}{2}} - \frac{1}{r + \frac{d}{2}} \right]\)
\(V = \frac{q}{4\pi\varepsilon_0 k} \left[ \frac{rd + \frac{d^2}{2} - (rd - \frac{d^2}{2})}{r^2 - (\frac{d}{2})^2} \right] = \frac{1}{4\pi\varepsilon_0 k} \frac{q \cdot d}{r^2 - \frac{d^2}{4}}\)
Since dipole moment \(p = q \cdot d\),
\(V = \frac{1}{4\pi\varepsilon_0 k} \frac{p}{r^2 - (\frac{d}{2})^2}\)
For a short dipole where \(d \ll r\), neglecting \(\left(\frac{d}{2}\right)^2\) in comparison to \(r^2\):
\(V = \frac{1}{4\pi\varepsilon_0 k} \frac{p}{r^2}\).

Teacher's Note:
a) Clearly define distances from both charges before applying the principle of superposition for scalar potentials.
b) State the short-dipole approximation clearly at the end.

14. Three capacitors o capacitance \(C_1 = 3\mu\text{F}\), \(C_2 = 6\mu\text{F}\) and \(C_3 = 10\mu\text{F}\) are connected to a 10v battery as shown in the figure below. [3 Marks]
(i) Equivalent
(ii) Electrostatic potential energy stored in the system.
[Figure: Circuit diagram showing capacitors C1 and C2 in series, combined in parallel with C3 across a 10V battery]

Answer:
(i) Capacitors \(C_1\) and \(C_2\) are in series combination. Their equivalent capacitance \(C'\) is:
\(\frac{1}{C'} = \frac{1}{3} + \frac{1}{6} = \frac{2 + 1}{6} = \frac{3}{6} = \frac{1}{2}\)
\(C' = 2\mu\text{F}\).
Now, \(C'\) and \(C_3\) (\(10\mu\text{F}\)) are in parallel combination. Total equivalent capacitance \(C = C' + C_3 = 2 + 10 = 12\mu\text{F}\).
(ii) Total charge \(Q = C \times V = 12\mu\text{F} \times 10\text{V} = 120\mu\text{C}\).
Electrostatic potential energy \(U = \frac{1}{2}QV = \frac{1}{2} \times 120 \times 10^{-6}\text{ C} \times 10\text{V} = 600 \times 10^{-6}\text{ J} = 6 \times 10^{-4}\text{ J}\).

Teacher's Note:
a) Solve series and parallel network combinations step-by-step from inner branches outward.
b) Ensure proper power-of-ten conversions for microfarads (\(\mu\text{F}\)) and microcoulombs (\(\mu\text{C}\)).

15. Draw a labelled circuit diagram of a potentiometer to measure the internal resistance r of a cell. Write the working formula (deviation is not required). [3 Marks]

Answer:
[Figure: Circuit diagram of a potentiometer setup with driver cell E', rheostat Rh, jockey J, galvanometer G, resistance box S, and cell E with plug keys K1 and K2]
Working formula derivation:
Let \(l_1\) be the balancing length when key \(K_2\) is open. Then \(E = kl_1\).
Let \(l_2\) be the balancing length when key \(K_2\) is closed with external resistance \(S\). Then terminal potential difference \(V = kl_2\).
Since \(\frac{E}{V} = \frac{l_1}{l_2}\) and internal resistance \(r = R \left(\frac{E}{V} - 1\right)\), the working formula is:
\(r = \left(\frac{l_1}{l_2} - 1\right)S\).

Teacher's Note:
a) Clearly label all components including the driver cell, auxiliary circuit, resistance box, and galvanometer.
b) State the final working formula clearly even if full algebraic derivation is skipped as per instructions.

16. A ray of light is incident on a prism whose refraction index is 1.52 at an angle of \(40^{\circ}\). If the angle of emergence is \(60^{\circ}\), calculation the angle of the prism. [3 Marks]

Answer:
Given \(n = 1.52\), angle of incidence \(i_1 = 40^{\circ}\), angle of emergence \(i_2 = 60^{\circ}\).
For the first refracting face (BC):
\(\sin r_1 = \frac{\sin i_1}{n} = \frac{\sin 40^{\circ}}{1.52} = \frac{0.6428}{1.52} \approx 0.4229\)
\(r_1 = \sin^{-1}(0.4229) \approx 25.01^{\circ}\).
For the second refracting face (BD):
\(\sin r_2 = \frac{\sin i_2}{n} = \frac{\sin 60^{\circ}}{1.52} = \frac{0.8660}{1.52} \approx 0.5697\)
\(r_2 = \sin^{-1}(0.5697) \approx 34.73^{\circ}\).
Angle of prism \(A = r_1 + r_2 = 25.01^{\circ} + 34.73^{\circ} \approx 59.74^{\circ}\) (or approx \(60^{\circ}\)).

Teacher's Note:
a) Apply Snell's law at each refracting surface independently to find internal angles \(r_1\) and \(r_2\).
b) The sum of the two internal angles equals the refracting angle of the prism (\(A = r_1 + r_2\)).

17. Derive the law of reflection using Huygens's wave theory. [3 Marks]

Answer:
[Figure: Ray diagram showing incident wavefront AB reflecting off a plane surface XY to form reflected wavefront CD, illustrating Huygens construction principles]
Let a plane wave front AB be incident obliquely on a reflecting surface XY. Let \(c\) be the speed of light.
In time \(t\), disturbance from B reaches C, travelling distance \(BC = ct\). Meanwhile, secondary waves from A emit a wavelet of radius \(AD = ct\) forming reflecting surface CD.
Consider triangles \(\Delta ABC\) and \(\Delta ADC\):
\(BC = AD = ct\) (radii of wavelets)
\(AC = AC\) (common hypotenuse)
\(\angle ABC = \angle ADC = 90^{\circ}\)
Therefore, \(\Delta ABC \cong \Delta ADC\) by RHS congruence.
Hence, \(\angle BAC = \angle DCA\), which means angle of incidence \(i\) equals angle of reflection \(r\) (\(\angle i = \angle r\)).
This proves the first law of reflection.

Teacher's Note:
a) Use congruent triangle geometry between incident and reflected wavefront constructions.
b) Clearly state that incident ray, normal, and reflected ray all lie in the same plane as the second law of reflection.

18. State any two Bohr's postulates and write the energy value of the ground state of the hydrogen atom. [3 Marks]

Answer:
1. Quantization condition: Electrons revolve only in certain stable, non-radiating circular orbits called stationary orbits where their orbital angular momentum is an integral multiple of \(\frac{h}{2\pi}\), i.e., \(mvr = \frac{nh}{2\pi}\).
2. Frequency condition: An electron can transition from a higher stationary orbit to a lower orbit by emitting a photon whose energy equals the energy difference between the two levels (\(\Delta E = h\nu\)).
Ground state energy of hydrogen atom: \(E = -13.6\text{ eV}\).

Teacher's Note:
a) State Bohr's angular momentum quantization condition clearly with mathematical expression.
b) Remember the negative sign in the ground state energy value representing bound state electron energy.

19. With reference to semiconductor answer the following. [3 Marks]
(i) What is the change in the resistance of the semiconductor with increase in temperature?
(ii) Name the majority charge carries in n-type semiconductor.
(iii) What is meant by doping?

Answer:
(i) The resistance (and resistivity) of a semiconductor decreases with an increase in temperature.
(ii) Electrons are the majority charge carriers in an n-type semiconductor.
(iii) Doping is the controlled process of adding impurity atoms to an intrinsic semiconductor to enhance its electrical conductivity.

Teacher's Note:
a) Semiconductors possess a negative temperature coefficient of resistance.
b) Pentavalent impurities are used for n-type doping while trivalent impurities are used for p-type doping.

SECTION D

20. (i) An alternating emf of 200V, 50Hz is applied to an L-R circuit, having a resistance R of \(10\Omega\) and an inductance L of \(0.05\text{H}\) connected in series. Calculate. [3 Marks]
(a) Impedance
(b) Current flowing in the circuit.

Answer:
Given \(V_{\text{rms}} = 200\text{ V}\), \(f = 50\text{ Hz}\), \(R = 10\Omega\), \(L = 0.05\text{ H}\).
Angular frequency \(\omega = 2\pi f = 2 \times \pi \times 50 = 100\pi\text{ rad/s}\).
Inductive reactance \(X_L = \omega L = 100\pi \times 0.05 = 5\pi\Omega \approx 5 \times 3.1416 = 15.71\Omega\).
(a) Impedance \(Z = \sqrt{R^2 + X_L^2} = \sqrt{10^2 + (15.71)^2} = \sqrt{100 + 246.8} = \sqrt{346.8} \approx 18.62\Omega\).
(b) Current flowing in the circuit \(I_{\text{rms}} = \frac{V_{\text{rms}}}{Z} = \frac{200}{18.62} \approx 10.74\text{ A}\).

Teacher's Note:
a) Calculate inductive reactance first using \(X_L = 2\pi fL\).
b) Impedance in an L-R series circuit is vectorially combined as \(Z = \sqrt{R^2 + X_L^2}\).

(ii) Draw a labelled graph showing the variation of inductive reactance \(X_L\) versus frequency f. [2 Marks]

Answer:
[Figure: Graph showing a straight line passing through the origin representing direct proportionality between inductive reactance \(X_L = 2\pi f L\) and frequency \(f\)]
Description of graph: A straight line starting from the origin passing upwards, indicating that inductive reactance is directly proportional to frequency (\(X_L \propto f\)).

Teacher's Note:
a) Clearly label axes with \(X_L\) on the vertical axis and frequency \(f\) on the horizontal axis.
b) The linear upward slope confirms that inductors block high-frequency AC more effectively.

21. Draw a neat labelled ray diagram showing the formation of an image at the least distance of distinct vision D by a simple microscope. When the final image is at D, derive an expansions for its magnifying power at D. [5 Marks]

Answer:
[Figure: Ray diagram of a simple microscope (convex lens) forming a virtual, magnified image at near point D from an object placed within focal length]
Derivation of magnifying power:
Angular magnification is defined as \(m = \frac{\beta}{\alpha}\), where \(\beta\) is the angle subtended by the image at the eye and \(\alpha\) is the angle subtended by the object at the unaided eye when both are placed at the least distance of distinct vision \(D\).
For small angles, \(\tan\alpha \approx \alpha\) and \(\tan\beta \approx \beta\).
From geometry, \(\tan\beta = \frac{A'B'}{D}\) and \(\tan\alpha = \frac{AB}{D}\).
Therefore, \(m = \frac{A'B' / D}{AB / D} = \frac{A'B'}{AB} = m'\) (linear magnification of lens).
Using lens formula for thin lens: \(\frac{1}{v} - \frac{1}{u} = \frac{1}{f}\).
Multiplying throughout by \(v\): \(1 - \frac{v}{u} = \frac{v}{f}\).
Substituting sign convention values (\(v = -D\)):
\(m = 1 - \frac{-D}{f} = 1 + \frac{D}{f}\).

Teacher's Note:
a) Ensure ray arrows and principal focus points F are clearly marked in the diagram.
b) State the final magnifying power formula clearly as \(m = 1 + \frac{D}{f}\) for a final image formed at the near point.

22. (I) Draw a labelled circuit diagram of a half-wave rectifier and give its output waveform. [3 Marks]

Answer:
[Figure: Circuit diagram of a half-wave rectifier showing step-down transformer, p-n junction diode in series, load resistor R_L, and input/output AC waveforms]
Working principle: During the positive half-cycle of the input AC, the diode is forward biased and conducts current, producing an output voltage across load resistor \(R_L\). During the negative half-cycle, the diode is reverse biased and blocks current flow, resulting in zero output voltage. Thus, only half-cycles are rectified.

Teacher's Note:
a) Label transformer windings, diode terminals (p and n), load resistor, and output terminals clearly.
b) Illustrate the pulsating unidirectional output wave corresponding to input alternating cycles.

(II) Draw a symbol for NOR gate and write its truth table. [2 Marks]

Answer:
[Figure: Logic symbol of a NOR gate showing OR gate combined with a NOT bubble at the output, with inputs A, B and output Y]
Truth table for NOR gate:

Teacher's Note:
a) A NOR gate is formed by connecting the output of an OR gate to the input of a NOT gate.
b) The output is high (1) only when all inputs are low (0).